Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n 2 1, 1 + 6 + 11 + 16 + ... + (5n - 4) - n(5n - 3) Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + ... + (5n - 4) = n(5n - 3) 2 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. O1=1: (5. 1- 3) 2 O P(1) = 1: (5 -1– 3) 2 O P(1) = 5 - 1 - 4 O1+ (5-1- 4) = 1. (5 - 1 – 3) The selected statement is true because both sides of the equation equal Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 1, and suppose that P(k) is true. The left-hand side of P(k) is -Select- , and the right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ** + (5(k + 1) - 4) = Hence P(k + 1) is true, which completes the inductive step. Select-- v + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It
Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n 2 1, 1 + 6 + 11 + 16 + ... + (5n - 4) - n(5n - 3) Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + ... + (5n - 4) = n(5n - 3) 2 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. O1=1: (5. 1- 3) 2 O P(1) = 1: (5 -1– 3) 2 O P(1) = 5 - 1 - 4 O1+ (5-1- 4) = 1. (5 - 1 – 3) The selected statement is true because both sides of the equation equal Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 1, and suppose that P(k) is true. The left-hand side of P(k) is -Select- , and the right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ** + (5(k + 1) - 4) = Hence P(k + 1) is true, which completes the inductive step. Select-- v + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
![Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2.
For every integer n 2 1, 1 + 6 + 11 + 16 + ... + (5n - 4) n(5n - 3)
Proof (by mathematical induction): Let P(n) be the equation
1 + 6 + 11 + 16 +... + (5n - 4) =
n(5n - 3)
2
We will show that P(n) is true for every integer n> 1.
Show that P(1) is true: Select P(1) from the choices below.
O1=1: (5 - 1 - 3)
2
O P(1) = 1: (5 ·1 – 3)
2
O P(1) = 5 - 1 - 4
O1+ (5 - 1 – 4) = 1 · (5·1 – 3)
The selected statement is true because both sides of the equation equal
Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k 2 1, and suppose that P(k) is true. The left-hand side of P(k) is -Select-
, and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + - + (5(k + 1) - 4) =
Hence P(k + 1) is true, which completes the inductive step.
Select-
v + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Need Help?
Read It](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99894a4a-cfa0-4bf4-9590-22770e9dd01d%2F6b3acf15-db4e-49d1-ab79-a8f34f005255%2Fkhru4ln_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2.
For every integer n 2 1, 1 + 6 + 11 + 16 + ... + (5n - 4) n(5n - 3)
Proof (by mathematical induction): Let P(n) be the equation
1 + 6 + 11 + 16 +... + (5n - 4) =
n(5n - 3)
2
We will show that P(n) is true for every integer n> 1.
Show that P(1) is true: Select P(1) from the choices below.
O1=1: (5 - 1 - 3)
2
O P(1) = 1: (5 ·1 – 3)
2
O P(1) = 5 - 1 - 4
O1+ (5 - 1 – 4) = 1 · (5·1 – 3)
The selected statement is true because both sides of the equation equal
Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k 2 1, and suppose that P(k) is true. The left-hand side of P(k) is -Select-
, and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + - + (5(k + 1) - 4) =
Hence P(k + 1) is true, which completes the inductive step.
Select-
v + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Need Help?
Read It
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
